# Applying Kepler's Laws

**Kepler's laws** describe the orbits of planets around the sun or stars around a galaxy in classical mechanics. They have been used to predict the orbits of many objects such as asteroids and comets , and were pivotal in the discovery of dark matter in the Milky Way. Violations of Kepler's laws have been used to explore more sophisticated models of gravity, such as general relativity. While Newton's laws generalize Kepler's laws, most problems related to the periods of orbits are still best solved using Kepler's laws, since they are simpler.

Recall the statements of Kepler's laws:

- Planets move in elliptical orbits with the sun at one focus.
- The line joining planets to either focus sweeps out equal areas in equal times.
- The square of the period is proportional to the cube of the semi-major axis (half the longer side of the ellipse): $T^2 \propto a^3.$

These laws can be applied to model natural objects like planets, stars, or comets, as well as man-made devices like rockets and satellites in orbit.

Although Kepler originally developed his laws in the context of planetary orbits, the results hold for any system with a radial force obeying the inverse square law. Coulomb's law holds that the electric force between two charged particles in an inverse square law like gravity (assuming that the particles have opposite charge). In spite of the fact that quantum mechanics is needed to fully model how electrons orbit nuclei, electrons with very high energy behave as though they had Keplerian orbits, and atoms containing such electrons are known as **Rydberg atoms**.

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## Halley's Comet

Halley's Comet is the first comet that astronomers realized had a periodic orbit. It passes within sight of Earth once every 75 years. Kepler's third law determines the length of the semimajor axis of this orbit:

$\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ \Rightarrow a&= \sqrt[3]{\frac{G M T^2}{4 \pi^2}}\\ &= 2.7 \times 10^{12} \text{ (m)}. \end{aligned}$

Above, the more precise form of Kepler's third law $T^2 = \frac{4 \pi^2}{G M} a^3$ has been used, where the proportionality constant between $T^2$ and $a^3$ has been solved for. Obtaining this constant requires an extensive derivation.

At the point in its orbit where it is closest to the sun, Halley's comet is only $8.8 \times 10^{8} \text{ m}$ from the sun, coming between the orbits of Mercury and Venus. It is furthest away from the sun at a distance of approximately $2a = 5.4 \times 10^{12} \text{ m}$, past the orbit of Neptune.

## Most Efficient Route to Mars

The most efficient route from Earth to Mars is called the Hohmann transfer orbit [2]. By Kepler's third law, it must be the ellipse that touches both orbits with the shortest possible semimajor axis. This is intuitive because without any acceleration, a rocket on Earth would stay in the orbit of Earth. The most efficient route is just the (elliptical) orbit which starts at the Earth's orbit and ends at the Mars' orbit.

The orbits of Earth and Mars are approximately circular, with radii $r_{E} = 1 \text{ AU}$ and $r_{M} = 1.542 \text{ AU}$. An ellipse touching both of these circles will have semimajor axis

$a = \frac{1}{2} (r_{E} + r_{M}) = 1.262 \text{ AU}.$

When $a$ is in units of $\text{AU}$ and $T$ is in units of years, Kepler's third law simplifies to the expression

$T^{2} = a^{3} \implies T = \sqrt{2.0992} = 1.41 \text{ (yr)}.$

## Straight Lines are Degenerate Ellipses

An astronaut stranded out in space wants to figure out how far she is from Earth. The only thing that she has with her is her handy dandy indestructible stopwatch that can survive the heat of re-entry and the shock of impact with Earth. She starts the stopwatch and lets it go with approximately zero velocity in the reference frame of Earth. After the watch crashes to Earth, mission control retrieves it and finds that the stopwatch was falling for 7 days. How far away from Earth is the astronaut?

This problem can be solved by integrating the Newtonian gravitational force over a straight line going towards Earth to find the position as a function of time and vice versa, but using the trick of applying Kepler's third law is easier. Suppose the astronaut's distance from Earth is some number $R$, and consider an ellipse with major axis of length slightly larger than $R$, with a minor axis $b \ll R$. This is an ellipse with eccentricity very near one. Taking the limit of this ellipse as the eccentricity goes to one yields a straight line of length R. This suggests that the properties of ellipses may be used to derive quantities related to straight lines in general, since any straight line may be approximated well by an ellipse of eccentricity near one.

The time $t$ it takes to fall down this line is half of the period $T$ it would take to go around the corresponding near-unit-eccentricity ellipse. Since the semimajor axis of the ellipse is $\frac{R}{2}$, by Kepler's third law

$\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ (2t)^2 &= \frac{4 \pi^2}{G M} \left(\frac{R}{2}\right)^3\\ \Rightarrow R &= 2 \sqrt[3]{\frac{G M t^{2}}{\pi^2}}. \end{aligned}$

Plugging in $G = 6.67 \times 10^{-11} \frac{\text{m}^3}{\text{ kg} \ \text{s}^2}$, $M = 5.97 \times 10^{24} \text{ kg}$, and $t = 6.048 \times 10^{5} \text{ s}$ gives

$R = 3.4 \times 10^{10} \text{ m}$

for this astronaut's distance from Earth. This derivation was much more straightforward than the burdensome calculus required to solve Newton's second law for the gravitational force.

## Deriving Velocity of Circular motion

Astronomers define a **galaxy rotation curve** to be the plot of the tangential velocity of stars as a function of their distance from the center of any given galaxy. Kepler's laws can be used to derive the shape of this curve assuming the stars have circular (or low eccentricity) orbits.

By Kepler's second law, the area swept out by the line from the galaxy center to a star in a given time must be constant. Since the radius $r$ to such a star is constant in circular motion, the velocity $v$ of these stars in orbit must also be constant. In that case, the period of rotation $T$ is $\frac{2 \pi r}{v}$. Plugging this into Kepler's third law gives

$\begin{aligned} T^2 &= \frac{4 \pi^2}{G M} a^3\\ \left(\frac{2 \pi r}{v}\right)^{2} &= \frac{\pi^2}{G M} r^{3}\\ v^{2} &= \frac{G M}{r}\\ \Rightarrow v(r) &= \sqrt{\frac{G M}{r}}. \end{aligned}$

The mass $M$ is also a function of the radius $r$, since even if the mass density in a galaxy is homogeneous this means that the mass contained in some radius scales as the radius cubed. That is, a cluster of stars with roughly homogeneous mass density will have mass that scales like $M(r) = M_{0} r^{3}$. Plugging this into the above expression for $v$ gives $v(r) = r \sqrt{G M_{0}}$.

A galaxy with a cluster of stars in the middle and sparsely spaced stars further out will therefore have a galaxy rotation curve that starts out linear and then falls off like $\frac{1}{\sqrt{r}}$, like curve A in the picture.

Astronomers have found that the galaxy rotation curve of the Milky Way does *not* follow this curve if we take into account all of the known mass in the Milky Way. It looks more like curve B, where the velocity stays approximately constant even far away from the center of the galaxy. This implies that either Kepler's laws are wrong or there is mass we haven't accounted for. While there are cases in which it is known that Kepler's laws are incomplete or not a good description of nature (see below), physicists are very confident that Kepler's laws should apply to these stars because their dynamics satisfy the relevant assumptions for Newtonian gravitation. The data predicts the existence of dark matter: matter that we can't see, but must be there due to its influence on gravitation.

## Violating Kepler's Laws

One of the greatest mysteries in early twentieth-century astronomy was the precession of the perihelion of orbit of the planet Mercury. The **perihelion** is the point in a planet's orbit at which it is closest to the sun. Kepler's first law predicts that the perihelion is constant in time for every orbit. However, the perihelion of Mercury moves every year by a small angle. Even accounting for perturbations due to the gravitational effects of other planets, there was still an unaccounted-for rotation of 43 arc-seconds per century when the orbit was solved in Newtonian gravity. The size of this effect is equivalent to a full rotation of the perihelion in 3 million years [4]. This slight precession of the orbit is an effect of general relativity, in which not all orbits are closed ellipses. Einstein's correct computation of the precession of the perihelion of Mercury in general relativity was hailed as one of the great early theoretical successes for the theory.

## Citations

[1] Image retrieved from http://apod.nasa.gov/apod/ap100104.html on 26 Feb 2016.

[2] Stern, David. *Kepler's Three Laws of Planetary Motion*. NASA. 23 Mar 2005. Retrieved on 9 Mar 2016 from http://www-spof.gsfc.nasa.gov/stargaze/Kep3laws.htm

[3] Image retrieved from https://commons.wikimedia.org/wiki/File:Mercure_orbite_precession_sk.JPG on 26 Feb 2016.

[4] Wudka, J. Precession of the perihelion of mercury. 24 Sep 1998. http://physics.ucr.edu/~wudka/Physics7/Notes_www/node98.html Retrieved on 18 Feb 2016.

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